Triangle parameterisation

Question

I get how to answer the qs below, the problem is actually finding path $2$ $ \left( 2, 0, 0 \right) $ to $ \left( 0, 1, 0 \right) $ I get $(2-t)i+tj$

yet the answer for path 2 is...

$$ (2-t)i+(t/2)j $$

Don't understand why, any help would be appreciated.

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Some-more context...

Let $G$ be the vector field given by

$$ G = 2{y i} + x^{2}{j} + z k $$

Evaluate the line integral

$$ I = \oint_{c} G \cdot dr $$

where $C$ is given by the three sides of the triangle with verticies $ \left( 0, 0, 0 \right) $, $ \left( 2, 0, 0 \right) $ and $ \left( 0, 0, 0 \right) $, and the integration is preformed in the following direction: from $ \left( 0, 0, 0 \right) $ to $ \left( 2, 0, 0 \right) $ then to $ \left( 0, 1, 0 \right) $ and finally back to $ \left( 0, 0, 0 \right) $. You may evaluate the integral $I$...

Answer 1

A parametric equation for a line, whose 2 points $\overrightarrow a$ and $\overrightarrow b$ are given, is - $$\overrightarrow r = \overrightarrow a + (\overrightarrow b - \overrightarrow a)k $$

Putting in the 2 points given, we get (writing in coordinate form) - \begin{align} \overrightarrow r &= (2,0,0) + ((0,1,0)-(2,0,0))k \\ &= (2,0,0) + (-2,1,0)k \end{align} Rescaling $ k = \frac{t}{2}$ to match with your values, we get - $$\overrightarrow r = (2-t)\hat i + \frac{t}{2}\hat j$$

Answer 2

The parametrisation of the line between two points

$$ A = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}, B = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}$$

is given by

$$\gamma(t) = A + (B - A)t$$

although other parametrisations are possible.

In particular, in your case, this yields

$$\gamma(t) = \begin{pmatrix} 2 \\ 0 \\0 \end{pmatrix} + (\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} - \begin{pmatrix} 2 \\ 0 \\ 0\end{pmatrix})t = \begin{pmatrix} 2 - 2t \\ t \\ 0 \end{pmatrix}$$

which is equivalent with your second parametrisation.

The main problem with your first parametrisation is that your second point does not lie on the path. You likely intended this to happen at $t = 1$; however, try entering that value in your parametrisation.